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1 year ago

Written as a rate, 100 miles in 5 hours would be ____ miles per hour.

Mathematics

1 answer:

Keith_Richards [23]1 year ago

4 0

Answer:

it would be 20 miles per hour

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Find the product -2X²×4X²y²

nydimaria [60]

Answer:

-8X^4y^2

Step-by-step explanation:

-2X² × 4X²y² = -8X^4y^2

edit to add:

first: multiply the coefficients together:

-2 x 4 = -8

second: add the exponents of X together (when you multiply exponents, you just add them together):

² & ² = ^4

and you have the answer:

-8X^4y^2

6 0

2 years ago

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I need an answer for question 16. I currently have 24 + 4 =

EastWind [94]

Answer:

Yes you are on the right track!

Step-by-step explanation:

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2 years ago

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What is -0.4 as a fraction or mixed number?<br><br>

NISA [10]

Answer:

-0.4 = −2/5

Step-by-step explanation:

Rewrite the decimal number as a fraction.

-0.4/1

Multiply to remove 1 decimal places.

-0.4/1 x 10/10 =4/10

Find the GCF.

4 ÷ 2/10 ÷ 2 = 2/5

x = 2/5

-2/5

3 0

2 years ago

A STERO is was priced for $75. If this is a 15% increase was is the new price?

Delicious77 [7]

Answer:

75+75×15/100=75+11.25=86.25$

7 0

2 years ago

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Listed below are speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, California (based on data from SigAlert

adell [148]

Answer:

Option (B) is the correct answer to the following question.

Step-by-step explanation:

Step-1: We have to find the Mean of the series.

The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.

$Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}$ $= 60.67$

Step-2: We have to find the Standard Deviation.

Let Standard Deviation be x.

Formula of Standard Deviation is: $s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}$

Put value in formula of Standard Deviation,

$s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}$ = 40.75

Step-3: Then, we have to find the critical value by chi-square.

$X_{1-\alpha/2}^{2}=3.82$

$X_{1-\alpha/2}^{2}=21.92$

Then, find the confidence interval which is 95%.

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9$

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9$

5 0

3 years ago